the Hölder $\rho$-continuity is defined as $$|f(x)-f(y)| \leq K_1 |x-y|^\rho.$$ I'm doing a research problem right now and might need the following condition $$K_2 |x-y|^\rho\leq|f(x)-f(y)| \leq K_1 |x-y|^\rho,$$ where both $K_1$ and $K_2$ are constants.
My question is, is it a high requirement to assume the left hand side of above condition?
If $\rho<1$ there are no such functions!
Assumme $f:\Bbb R\to \Bbb R$ and $|x-y|^{1/2}\le|f(x)-f(y)|$. Say $f(0)=0$. Now for every $x>0$ either $f(x)\ge x^{1/2}$ or $f(x)\le-x^{1/2}$. By continuity, wlog $$f(x)\ge x^{1/2}\quad(x>0).$$
Fix $x>0$ for a second. If we had $f(x+y)\le f(x)-y^{1/2}$ for every $y>0$ then large $y$ would contradict the previous inequality. So $$f(x+y)\ge x^{1/2}+y^{1/2}\quad(x,y>0).$$By induction $$f\left(\sum_{j=1}^nx_j\right)\ge\sum_{j=1}^nx_j^{1/2}\quad(x_j>0).$$Now $1=\sum_{j=1}^n1/n$ implies that $f(1)=\infty$.
Similarly there is no $f:[0,1]\to\Bbb R$ such that $|x-y|^{1/2}\le|f(x)-f(y)|\le c|x-y|^{1/2}$. As before we can assume that $f(x)\ge x^{1/2}$. Now above we needed large $y$ to get a contradiction from $f(x+y)\le f(x)-y^{1/2}$. Under our current assummptions this inequality implies $f(x+y)\le cx^{1/2}-y^{1/2}$, giving a bound on how large $y$ has to be to get that contradiction; if $x>0$ is small enough then we get a contradiction from $y=1-x$. If I did the sums correctly it seems to me that we have $$f(x+y)\ge x^{1/2}+y^{1/2}\quad(x,y>0,x+y<\delta)$$where $\delta=1/(1+(c-1)^2)$. This leads to $f(\delta)=\infty$ as before.
Your question is very vague and depending on the context, you will get different answers. I think it is reasonable to make such an assumption to derive a technical result. At the very least, it is a starting point towards understanding and deriving properties of the problem you are working on. Later, you can try relaxing this assumption.
For what it is worth, I will mention one well-known example where such a condition is imposed in the theoretical analysis of an algorithm, namely the convergence of gradient descent for convex, differentiable functions.
In particular assume that you want to minimize $f(x)$, $x \in \mathbb R^n$, where $f$ is convex, differentiable function. One possibility is to use the gradient descent algorithm. Under suitable choices of the step-size, the algorithm then can be shown to converge at a linear rate, if the gradient of $f$ is Lipschitz with constant L>0, i.e.
$$\vert\vert \nabla f(x) - \nabla f(y) \vert \vert_2 \leq L \vert\vert x-y\vert\vert_2$$
OTOH, the convergence can be shown to be exponentially fast, when $f$ is also strongly convex with modulus $m>0$ ($m<L$). This is equivalent to the following condition:
$$m \vert\vert x-y\vert\vert_2 \leq \vert\vert \nabla f(x) - \nabla f(y) \vert \vert_2 \leq L \vert\vert x-y\vert\vert_2$$
So this is an example where a requirement like the one you stated is used in the theoretical analysis of an algorithm.
